Construction: I draw a perpendicular AD on the hypotenuse BC from the right angular point A which intersects the side BC at the point D.

Proof: In right-angled Δ ABC, AD is perpendicular on the hypotenuse BC.

∴ Δ ABD and Δ CBA are similar.

Hence,

{\text{AB} \over \text{BC}} = {\text{BD} \over \text{AB}},

∴ AB² = BC × BD —- (i)

Again, ΔCAD and ΔCBA are similar.

Hence,

{\text{AC} \over \text{BC}} = {\text{DC} \over \text{AC}}

∴ AC² = BC × DC —- (ii)

So, by adding (i) and (ii), I get:

AB² + AC² = BC × BD + BC × DC.

= BC (BD + DC) = BC × BC = BC².

∴ BC² = AB² + AC² [proved]